Question

In Problems 49–52, find the area of the parallelogram with vertices $P_1,P_2,P_3,P_4$

role="math" localid="1646760204418" $$\begin{gathered} P_1=\left(2,1,1\right) \\ {P}_2=\left(2,3,1\right) \\ {P}_3=\left(-2,4,1\right) \\ {P}_4=\left(-2,6,1\right) \end{gathered}$$

Short answer

The area of the parallelogram is$8$ square units.

Step-by-step solution

Step 1. Given data

Vertices of the parallelogram are $$\begin{gathered} P_1=\left(2,1,1\right) \\ {P}_2=\left(2,3,1\right) \\ {P}_3=\left(-2,4,1\right) \\ {P}_4=\left(-2,6,1\right) \end{gathered}$$

The area of the Parallelogram is the magnitude of the cross product of the adjacent sides.

$A=\left|\left|u\times v\right|\right|$

Step 2. Finding adjacent sides

Let's say adjacent sides are $u,v$.

$$\begin{gathered} u=\stackrel{\rightharpoonup }{P_1{P}_2} \\ \Rightarrow u=0i+2j+0k \\ v=\stackrel{\rightharpoonup }{P_1{P}_3} \\ \Rightarrow v=-4i+3j+0k \end{gathered}$$

Step 3. Finding the area

The crossproduct:

$$\begin{gathered} u\times v=\left|\begin{array}{ccc}i& j& k\\ 0& 2& 0\\ -4& 3& 0\end{array}\right| \\ u\times v=\left|\begin{array}{cc}2& 0\\ 3& 0\end{array}\right|i-\left|\begin{array}{cc}0& 0\\ -4& 0\end{array}\right|j+\left|\begin{array}{cc}0& 2\\ -4& 3\end{array}\right|k \\ u\times v=8k \end{gathered}$$

Step 4. Finding the area

The area:

$$\begin{gathered} A=\left|\left|u\times v\right|\right| \\ A=\left|\left|8k\right|\right| \\ A=8 \end{gathered}$$

Therefore the area is$8$square units.